\newproblem{lay:6_4_19}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 6.4.19}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Suppose that $A=QR$ where $Q$ is $m\times n$ and $R$ is $n\times n$. Show that if the columns of $A$ are linearly independent, then $R$ mut be invertible.
	[\textit{Hint}: Study the equation $R\mathbf{x}=\mathbf{0}$ and use the fact that $A=QR$.]
}{
   % Solution
	Since $Q$ is orthogonal and meets that $Q^TQ=I$ we have
	\begin{center}
		$A=QR$ \\
		$Q^TA=R$ \\
	\end{center}
	Then, the equation $R\mathbf{x}=\mathbf{0}$ becomes
	\begin{center}
		$Q^TA\mathbf{x}=\mathbf{0}$
	\end{center}
	Multiplying both sides by $Q$ we have
	\begin{center}
		$A\mathbf{x}=Q\mathbf{0}=\mathbf{0}$
	\end{center}
	Since the columns of $A$ are linearly independent, the only solution of this problem is $\mathbf{x}=\mathbf{0}$ (see Equation 1.7.3) and, consequently, the only
	solution of $R\mathbf{x}=\mathbf{0}$ is also $\mathbf{x}=\mathbf{0}$. But this implies, by the Invertible Matrix Theorem, that $R$ is invertible.
}
\useproblem{lay:6_4_19}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
